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When a deuteron of mass 2.0141 a.m.u and negligible K.E. is absorbed by a Lithium `(._(3)Li^(6))` nucleus of mass 6.0155 a.m.u. the compound nucleus disintegration spontaneously into two alpha particles, each of mass 4.0026 a.m.u. Calculate the energy carried by each `alpha` particle. |
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Answer» `._(3)^(6)Ll+._(1)^(2)Drarr._(2)^(4)He+._(2)^(4)He+Q` `m(._(3)^(6)Li)=6.0155u` `m(._(1)^(2)H)=2.0141 u` Total initial mass `=8.0296 u` TOtal final mass `=2m(._(2)^(4)He)=2xx4.0026` `=8.0052u` Mass defect, `Delta m=8.296-8.0052` `=0.0244u=0.024xx1.66xx10^(-27)kg` Energy released, `Q-Deltamxxc^(2)=0.244xx1.66xx10^(27)xx(3xx10^(8))^(2)` `=3.645xx10^(-12)J` Energy of each `alpha`-particle=`1.8225xx10^(-12)J` |
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